SOLUTION: A cyclist rode the first 16 mile-portion of his workout at a constant speed. For the 12 mile cool down portion of his workout, he reduced his speed by 2 miles per hour. Each portio

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Question 581323: A cyclist rode the first 16 mile-portion of his workout at a constant speed. For the 12 mile cool down portion of his workout, he reduced his speed by 2 miles per hour. Each portion took the same time. Find his speed during the first portion and the second cool down portion.
Found 2 solutions by stanbon, richwmiller:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A cyclist rode the first 16 mile-portion of his workout at a constant speed. For the 12 mile cool down portion of his workout, he reduced his speed by 2 miles per hour. Each portion took the same time. Find his speed during the first portion and the second cool down portion.
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1st portion DATA:
distance = 16 miles ; rate = x mph ; time = d/r = 16/x hrs
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2nd portion DATA:
distance = 12 miles ; rate = (x-2)mph ; time = 12/(x-2) hrs
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Equation:
time = time
16/x = 12/(x-2)
16x-32 = 12x
4x = 32
x = 8 mph (rate of 1st portion)
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x-2 = 6 mph (rate on 2nd portion)
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Cheers,
Stan H.

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
16+12=28
r*t=d=16
R*T=D=12
t=T
r*t=16
R*t=12
R=r-2
(r-2)*t=12
t=12/(r-2)
16/r=12/(r-2)
12r=16(r-2)
12r=16r-32
0=4r-32
32=4r
8=r
6=R